Peter‘s Hobby

Problem Description

Recently, Peter likes to measure the humidity of leaves. He recorded a leaf humidity every day. There are four types of leaves wetness: Dry , Dryish , Damp and Soggy. As we know, the humidity of leaves is affected by the weather. And there are only three kinds of weather: Sunny, Cloudy and Rainy.For example, under Sunny conditions, the possibility of leaves are dry is 0.6.
Give you the possibility list of weather to the humidity of leaves.


The weather today is affected by the weather yesterday. For example, if yesterday is Sunny, the possibility of today cloudy is 0.375.
The relationship between weather today and weather yesterday is following by table:


Now,Peter has some recodes of the humidity of leaves in N days.And we know the weather conditons on the first day : the probability of sunny is 0.63,the probability of cloudy is 0.17,the probability of rainny is 0.2.Could you know the weathers of these days most probably like in order?

 

Input

The first line is T, means the number of cases, then the followings are T cases. for each case:
The first line is a integer n(n<=50),means the number of days, and the next n lines, each line is a string shows the humidity of leaves (Dry, Dryish, Damp, Soggy)

 

Output

For each test case, print the case number on its own line. Then is the most possible weather sequence.( We guarantee that the data has a unique solution)

 

Sample Input

1
3
Dry
Damp
Soggy

 

Sample Output

Case #1:
Sunny
Cloudy
Rainy

Hint
Log is useful.
 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define maxn 105
#define MAXN 100005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-10
typedef long long ll;
using namespace std;

int n,m,flag,cnt,tot,test=0;
int num[55],pre[55][4];
double dp[55][4];
char s[50];
double lea[3][4]=
{
    {0.6, 0.2, 0.15, 0.05},
    {0.25, 0.3, 0.2, 0.25},
    {0.05, 0.10, 0.35, 0.50}
};
double wea[3][3]=
{
    {0.5, 0.375, 0.125},
    {0.25, 0.125, 0.625},
    {0.25, 0.375, 0.375}
};
char res[3][15]=
{
    "Sunny","Cloudy","Rainy"
};

void output(int x,int y)
{
    if(x==0) return ;
    output(x-1,pre[x][y]);
    printf("%s\n",res[y]);
}
void solve()
{
    int i,j,k,t,id;
    double ma,tmp;
    dp[1][0]=0.63*lea[0][num[1]];
    dp[1][1]=0.17*lea[1][num[1]];
    dp[1][2]=0.2*lea[2][num[1]];
    for(i=2;i<=n;i++)
    {
        for(j=0;j<3;j++)
        {
            ma=-1;
            for(k=0;k<3;k++)
            {
                tmp=dp[i-1][k]*wea[k][j]*lea[j][num[i]];
                if(ma<tmp)
                {
                    ma=tmp; id=k;
                }
            }
            dp[i][j]=ma; pre[i][j]=id;
        }
    }
    ma=-1;
    for(j=0;j<3;j++)
    {
        if(dp[n][j]>ma)
        {
            ma=dp[n][j]; id=j;
        }
    }
    printf("Case #%d:\n",++test);
    output(n,id);
}
int main()
{
    int i,j,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%s",s);
            if(strcmp(s,"Dry")==0) num[i]=0;
            else if(strcmp(s,"Dryish")==0) num[i]=1;
            else if(strcmp(s,"Damp")==0) num[i]=2;
            else num[i]=3;
        }
        solve();
    }
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define maxn 105
#define MAXN 100005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-10
typedef long long ll;
using namespace std;

int n,m,flag,cnt,tot,test=0;
int num[55],pre[55][4];
double dp[55][4];
char s[50];
double lea[3][4]=
{
    {0.6, 0.2, 0.15, 0.05},
    {0.25, 0.3, 0.2, 0.25},
    {0.05, 0.10, 0.35, 0.50}
};
double wea[3][3]=
{
    {0.5, 0.375, 0.125},
    {0.25, 0.125, 0.625},
    {0.25, 0.375, 0.375}
};
char res[3][15]=
{
    "Sunny","Cloudy","Rainy"
};

void output(int x,int y)
{
    if(x==0) return ;
    output(x-1,pre[x][y]);
    printf("%s\n",res[y]);
}
void solve()
{
    int i,j,k,t,id;
    double ma,tmp;
    dp[1][0]=log(0.63)+lea[0][num[1]];
    dp[1][1]=log(0.17)+lea[1][num[1]];
    dp[1][2]=log(0.2)+lea[2][num[1]];
    for(i=2;i<=n;i++)
    {
        for(j=0;j<3;j++)
        {
            ma=-INF;
            for(k=0;k<3;k++)
            {
                tmp=dp[i-1][k]+wea[k][j]+lea[j][num[i]];
                if(ma<tmp)
                {
                    ma=tmp; id=k;
                }
            }
            dp[i][j]=ma; pre[i][j]=id;
        }
    }
    ma=-INF;
    for(j=0;j<3;j++)
    {
        if(dp[n][j]>ma)
        {
            ma=dp[n][j];
            id=j;
        }
    }
    printf("Case #%d:\n",++test);
    output(n,id);
}
int main()
{
    int i,j,t;
    for(i=0;i<3;i++)
    {
        for(j=0;j<4;j++)
        {
            lea[i][j]=log(lea[i][j]);
            if(j<3) wea[i][j]=log(wea[i][j]);
        }
    }
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%s",s);
            if(strcmp(s,"Dry")==0) num[i]=0;
            else if(strcmp(s,"Dryish")==0) num[i]=1;
            else if(strcmp(s,"Damp")==0) num[i]=2;
            else num[i]=3;
        }
        solve();
    }
    return 0;
}